Substituting y = x^4, we get

y + √(y+20) = 22

√(y+20) = 22 - y

y + 20 = (22 - y)^2 = 484 - 44y + y^2

y^2 - 45y + 464 = 0

(y - 29)(y - 16) = 0

x^4 = 16 has

X = 2, x = -2

It also has 6 other roots, of which 4 are imaginary, assuming the "under root" can be taken as positive or negative.

y + √(y+20) = 22

√(y+20) = 22 - y

y + 20 = (22 - y)^2 = 484 - 44y + y^2

y^2 - 45y + 464 = 0

(y - 29)(y - 16) = 0

x^4 = 16 has

**two integral roots**:X = 2, x = -2

It also has 6 other roots, of which 4 are imaginary, assuming the "under root" can be taken as positive or negative.